Determine the shortest distance from the line 12x+5y-3=0 to the point (1, -4). Round your answer to the nearest tenth.

 


To find the shortest distance from a point to a line, we can use the formula:

\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

Where:
- \( A, B, C \) are the coefficients from the line equation \( Ax + By + C = 0 \).
- \( (x_1, y_1) \) is the point.
- \( d \) is the distance.

The given line equation is \( 12x + 5y - 3 = 0 \), so:
- \( A = 12 \)
- \( B = 5 \)
- \( C = -3 \)

The given point is \( (1, -4) \), so \( x_1 = 1 \) and \( y_1 = -4 \).

Substituting these values into the formula:

\[
d = \frac{|12(1) + 5(-4) - 3|}{\sqrt{12^2 + 5^2}}
\]

Let's calculate it step by step.

1. The numerator:
\[
|12(1) + 5(-4) - 3| = |12 - 20 - 3| = |-11| = 11
\]

2. The denominator:
\[
\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13
\]

So, the distance is:

\[
d = \frac{11}{13} \approx 0.846
\]

Rounded to the nearest tenth, the shortest distance is **0.8**.

Would you like any additional details or explanations on this?

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Here are five related questions:
1. How do you find the distance between two points?
2. How do you derive the equation of a line given a point and slope?
3. What is the slope of the line \(12x + 5y - 3 = 0\)?
4. Can you generalize the formula for the distance from a point to a plane in 3D?
5. How does changing the coefficients in the line equation affect the distance?

**Tip**: Always verify whether your distance formula needs to be applied to normalized or standard line forms for accuracy.

Determine the shortest distance from the line 12x+5y-3=0 to the point (1, -4). Round your answer to the nearest tenth.

Discover how to determine the shortest distance from the line 12x + 5y - 3 = 0 to the point (1, -4) using the distance formula. Step-by-step calculation and clear explanation provided. Ideal for students in Grades 9-12.

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